CHAPTER 11 CHI-SQUARE AND F DISTRIBUTIONS

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1 CHAPTER 11 CHI-SQUARE AND F DISTRIBUTIONS CHI-SQUARE TESTS OF INDEPENDENCE (SECTION 11.1 OF UNDERSTANDABLE STATISTICS) In chi-square tests of independence we use the hypotheses. H0: The variables are independent H1: The variables are not independent To use MINITAB for tests of independence, we enter the values of a contingency table row by row. The command CHISQUARE then prints a contingency table showing both the observed and expected counts. It computes the sample chi-square value using the following formula, in which E stands for the expected count in a cell and O stands for the observed count in that same cell. The sum is taken over all cells. χ 2 2 ( O E) = E Then MINITAB gives the number of degrees of the chi-square distribution. To conclude the test, use the P value of the sample chi-square statistic if your version of MINITAB provides it. Otherwise, compare the calculated chi-square value to a table of the chi-square distribution with the indicated degrees of freedom. We may use Table 8 of Appendix II of Understandable Statistics. If the calculated sample chi-square value is larger than the value in Table 8 for a specified level of significance, we reject H 0. Use the menu selection Stat Tables Chi-square Test Dialog Box Responses Example List the columns containing the data from the contingency table. Each column must contain integer values. A computer programming aptitude test has been developed for high school seniors. The test designers claim that scores on the test are independent of the type of school the student attends: rural, suburban, urban. A study involving a random sample of students from these types of institutions yielded the following contingency table. Use the CHISQUARE command to compute the sample chi-square value, and to determine the degrees of freedom of the chi-square distribution. Then determine if type or school and test score are independent at the α = 0.05 level of significance. School Type Score Rural Suburban Urban To use the menu selection Stat Tables Chi-square Test with C1 containing test scores for rural schools, C2 corresponding test scores for suburban schools, and C3 corresponding test scores for urban schools. 267

2 268 Technology Guide Understandable Statistics, 8th Edition Since the P value, 0.855, is greater thanα = 0.05, we do not reject the null hypothesis. LAB ACTIVITIES FOR CHI-SQUARE TESTS OF INDEPENDENCE Use MINITAB to produce a contingency table and compute the sample chi square value. If your version of MINITAB produces the P value of the sample chi-square statistic, conclude the test using P values. Otherwise, use Table 9 of Understandable Statistics to find the chi-square value for the given α and degrees of freedom. Compare the sample chi-square value to the value found in Table 8 to conclude the test. 1. We Care Auto Insurance had its staff of actuaries conduct a study to see if vehicle type and loss claim are independent. A random sample of auto claims over six months gives the information in the contingency table. Total Loss Claims per Year per Vehicle Type of vehicle $0 999 $ $ $6000+ Sports car Truck Family Sedan Compact Test the claim that car type and loss claim are independent. Use α = 0.05.

3 Part III: MINITAB Guide An educational specialist is interested in comparing three methods of instruction. SL standard lecture with discussion TV video taped lectures with no discussion IM individualized method with reading assignments and tutoring, but no lectures. The specialist conducted a study of these methods to see if they are independent. A course was taught using each of the three methods and a standard final exam was given at the end. Students were put into the different method sections at random. The course type and test results are shown in the next contingency table. Final Exam Score Course Type < SL TV IM Test the claim that the instruction method and final exam test scores are independent, using α = ANALYSIS OF VARIANCE (ANOVA) (SECTION 11.5 OF UNDERSTANDABLE STATISTICS) Section 11.5 of Understandable Statistics introduces single factor analysis of variance (also called one-way ANOVA). We consider several populations that are each assumed to follow a normal distribution. The standard deviations of the populations are assumed to be approximately equal. ANOVA provides a method to compare several different populations to see if the means are the same. Let population 1 have mean µ 1, population 2 have mean µ 2, and so forth. The hypotheses of ANOVA are H0: u1 = u2 = K = un H1: not all the means are equal. In MINITAB we use the menu selection Stat ANOVA Oneway (Unstacked) to perform one-way ANOVA. We put the data from each population in a separate column. The different populations are called levels in the output of ANOVAONEWAY. An analysis of variance table is printed, as well a confidence interval for the mean of each level. If there are only two populations, ANOVAONEWAY is equivalent to using the 2-Sample Test choice with the equal variances option checked.

4 270 Technology Guide Understandable Statistics, 8th Edition Stat ANOVA Oneway (Unstacked) Dialog Box Responses Example Responses: Enter the columns containing the data. Select a confidence level such as 95%. Check [Store Residuals] and /or [Store fits] only when you want to store these results. A psychologist has developed a series of tests to measure a person s level of depression. The composite scores range from 50 to 100 with 100 representing the most severe depression level. A random sample of 12 patients with approximately the same depression level, as measured by the tests, was divided into 3 different treatment groups. Then, one month after treatment was completed, the depression level of each patient was again evaluated. The after-treatment depression levels are given below. Treatment Treatment Treatment Put treatment 1 responses in column C1, treatment 2 responses in C2, treatment 3 responses in C3. Use the Stat ANOVA Oneway (Unstacked) menu selections.

5 Part III: MINITAB Guide 271 The results are Since the level of significance α = 0.05 is less than the P value of 0.965, we do not reject H 0. LAB ACTIVITIES FOR ANALYSIS OF VARIANCE 1. A random sample of 20 overweight adults was randomly divided into 4 groups. Each group was given a different diet plan, and the weight loss for each individual after 3 months follows: Plan Plan Plan Plan Test the claim that the population mean weight loss is the same for the four diet plans, at the 5% level of significance. 2. A psychologist is studying the time it takes rats to respond to stimuli after being given doses of different tranquilizing drugs. A random sample of 18 rats was divided into 3 groups. Each group was given a different drug. The response time to stimuli was measured (in seconds). The results follow. Drug A Drug B Drug C

6 272 Technology Guide Understandable Statistics, 8th Edition Test the claim that the population mean response times for the three drugs is the same, at the 5% level of significance. 3. A research group is testing various chemical combinations designed to neutralize and buffer the effects of acid rain on lakes. Eighteen lakes of similar size in the same region have all been affected in the same way by acid rain. The lakes are divided into four groups and each group of lakes is sprayed with a different chemical combination. An acidity index is then take after treatment. The index ranges from 60 to 100, with 100 indicating the greatest acid rain pollution. The results follow. Combination I Combination II Combination III Test the claim that the population mean acidity index after each of the four treatments is the same at the 0.01 level of significance. COMMAND SUMMARY CHISQUARE C C produced a contingency table and computes the sample chi-square value WINDOWS menu select: Stat Tables Chi-square test In the dialog box, specify the columns that contain the chi-square table. AOVONEWAY C C performs a one-way analysis of variance. Each column contains data from a different population WINDOWS menu select: Stat ANOVA Oneway (Unstacked) In the dialog box specify the columns to be included.